our operator is just differentiation so, yes, . More generally, if , then is an "eigenvector" and is an eigenvalue. As for finding , you don't! That equation is true for all - any real number is an eigenvalue.

As for whether or not it is "self adjoint", the definition of self-adjoint is that <Af, g>= <f, Ag> for any vectors f, g. Here, I presume you are using the "standard" inner product, so that we want . Use "integration by parts" with u= g and dv= (df/dx)dx. Then du= (dg/dx)dx and v= f.

provided f(1)g(1)- f(0)g(0)= 0.